The Student Room Group

Geometric series proof [Edexcel C2]

10. A geometric series is a+ar+ar2+...a + ar + ar^2 + ...

a) Prove that the sum of the first nn terms of this series is given by

Sn=a(1rn)1rS_n = \frac{a(1-r^n)}{1-r}

[4 marks]
Obviously that's the formula they give you so I've always sort of taken it as a given. Thus I have no idea how to prove that it is true...can anyone help me out and guide me through how to answer this question?

Thanks. :smile:
Reply 1
sn=a+ar+ar2+...s_n=a+ar+ar^2+...

What's rsnrs_n and then what's snrsns_n-rs_n?

That should help you prove it.
Reply 2
Sn=a+ar+ar2........arn2+arn1S_n = a+ ar +ar^2........ar^{n-2} + ar^{n-1}

rSn=ar+ar2+ar3........arn1+arnrS_n = ar+ ar^2 +ar^3........ar^{n-1} + ar^n

SnrSn=aarnS_n-rS_n =a-ar^n

Sn(1r)=a(1rn)S_n(1-r) = a(1-r^n)

Sn=a(1rn)1rS_n= \frac{a(1-r^n)}{1-r}

Geddit?
Reply 3
Thanks!
Reply 4
Although 3 years is a while for a thanks
I looked at the answer and used a quotient of (1 - r) top and bottom which cancels a lot of terms on top. Not formal but it's nice!
Original post by IChem
Sn=a+ar+ar2........arn2+arn1S_n = a+ ar +ar^2........ar^{n-2} + ar^{n-1}

rSn=ar+ar2+ar3........arn1+arnrS_n = ar+ ar^2 +ar^3........ar^{n-1} + ar^n

SnrSn=aarnS_n-rS_n =a-ar^n

Sn(1r)=a(1rn)S_n(1-r) = a(1-r^n)

Sn=a(1rn)1rS_n= \frac{a(1-r^n)}{1-r}

Geddit?


7 years later and still useful :wink:

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